Stress, the Cauchy tensor and Mohr's circle
Traction vector and stress tensor, plane stress on a rotated plane, Mohr's circle proved, principal stresses and maximum shear with an example, 45° fractures.
Internal forces are resultants: to know whether the material resists, one must know how force is distributed point by point. The quantity describing it is stress, force per unit area. At a point, stress depends on the orientation of the face on which it is viewed. In this lesson we look at the stress vector and tensor, stresses on a rotated plane in the plane case, Mohr’s circle with its proof and an example, and principal stresses.
The stress tensor
On a small face of area 
and normal 
, inside a body, acts a force 
; the traction vector is 
, measured in pascals (in practice in N/mm² = MPa). It splits into a normal component 
, which pulls or compresses the face, and a shear component 
, which makes it slide.
Cauchy proved that knowing the stresses on three perpendicular faces is enough to obtain them on any face: the nine components form the stress tensor, which is symmetric (
, by rotational equilibrium of an elementary cube), and
![\[ \mathbf{t} = \boldsymbol{\sigma}\,\mathbf{n} \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-d3aa5f077dd0f6467bbe07aa11d6a617_l3.png "Rendered by QuickLaTeX.com")
Plane stress
In many cases (thin plates, free surfaces, beams) the stresses lie in a plane: 
, 
and 
. On a face whose normal makes angle 
with the 
axis, projecting 
onto the normal and the tangent and using the double-angle formulas:
![\[ \sigma_n = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-8aa80c7bee0ce75d282242c8a086f22c_l3.png "Rendered by QuickLaTeX.com")
![\[ \tau_n = -\frac{\sigma_x - \sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-572d3ee894958d6a21205016eee5e1fb_l3.png "Rendered by QuickLaTeX.com")
Mohr’s circle
Theorem. As the orientation varies, the point 
describes a circle with centre 
and radius
![\[ R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-ae4934dd5ec0521fcf5a1d3992963de5_l3.png "Rendered by QuickLaTeX.com")
Proof. Setting 
and 
, the previous formulas become 
and 
. Squaring and adding, the cross terms cancel and, since 
, what remains is 
: the equation of a circle. ∎
Rotating the face by in the body means rotating by 
on the circle. The faces with normals and 
are represented by two diametrically opposite points, X and Y.
Principal stresses
The intersections of the circle with the axis are the principal stresses:
![\[ \sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm R \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-a066c38bbe93bfb87261995cd9ad1a00_l3.png "Rendered by QuickLaTeX.com")
on whose faces the shear stress is zero. The maximum shear stress is the radius 
, on faces inclined at 45° to the principal directions.
Example. 
MPa, 
MPa, 
MPa. Centre 50, radius 
MPa. Principal stresses 
MPa and 
; maximum shear 50 MPa. The principal direction is found from 
, i.e. 
and 
.
Principal stresses explain many failures. A twisted piece of chalk breaks along a 45° helix: torsion produces pure shear, corresponding to principal tensile stresses at 45°, and chalk, being brittle, fails in tension. In reinforced concrete shear cracks are inclined for the same reason (Building Techniques lesson on reinforced concrete in shear). Strain-gauge measurements (Electrical Measurements lesson on sensors) are often made with rosettes of three grids at 0°, 45° and 90°, from which the plane state, and hence Mohr’s circle, is reconstructed.
